Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
long long L[MAXN], R[MAXN], S[MAXN];
const double eps = 1e-6, inf = 1e15;
int main() {
  long long sy, a, b;
  scanf("%lld%lld%lld", &sy, &a, &b);
  int N;
  scanf("%d", &N);
  L[0] = R[0] = 0;
  L[N + 1] = R[N + 1] = inf;
  for (int i = 1; i <= N; ++i) scanf("%lld%lld", &L[i], &R[i]);
  S[0] = 0;
  for (int i = 1; i <= N; ++i) S[i] = S[i - 1] + R[i] - L[i];
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    double x_a = a + double(x - a) / double(sy - y) * sy,
           x_b = b + double(x - b) / double(sy - y) * sy;
    int l = 0, r = N + 1;
    while (l != r) {
      int m = (l + r + 1) / 2;
      if (L[m] <= x_a + eps)
        l = m;
      else
        r = m - 1;
    }
    int i = l;
    l = 0, r = N + 1;
    while (l != r) {
      int m = (l + r) / 2;
      if (R[m] >= x_b - eps)
        r = m;
      else
        l = m + 1;
    }
    int j = r;
    double ans = 0.0;
    if (i == j)
      ans = x_b - x_a;
    else
      ans = S[j - 1] - S[i] + max(0.0, R[i] - x_a) + max(0.0, x_b - L[j]);
    printf("%.10f\n", ans * (double(y - sy) / y));
  }
  return 0;
}